In the given circuit, if the current flowing through 5 \( \Omega \) resistor is 0.5 A, then the value of \( E \) is:
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In complex resistor networks, use Ohm's law and equivalent resistance to simplify the circuit before applying Kirchhoff's laws or calculating voltages.
The current flowing through the 5 \( \Omega \) resistor is given as 0.5 A. Using Ohm's law, \( V = IR \), we can find the voltage across the 5 \( \Omega \) resistor.
\[
V = I \times R = 0.5 \, \text{A} \times 5 \, \Omega = 2.5 \, \text{V}
\]
Now, considering the series and parallel combinations of resistors, we can calculate the total voltage \( E \) applied to the circuit by adding the appropriate voltage contributions across the resistors. After doing the necessary calculations for each resistor in the network, we find the value of \( E \) to be 6 V.
Thus, the correct answer is option (2), 6 V.
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Approach Solution -2
In the given circuit, if the current flowing through 5 Ω resistor is 0.5 A, then the value of E is:
Given: Current through 5 Ω resistor = 0.5 A
Using Ohm’s law: V = I × R
Voltage drop across 5 Ω = 0.5 × 5 = 2.5 V
Now trace the entire loop including the 5 Ω resistor and the EMF source E.
We need to find total potential drop across this loop.
In the loop from battery E (5 Ω in series with two 10 Ω in parallel and then 8 Ω, 3 Ω, and 6 Ω connected in a complex parallel-series),
effective voltage drop across the rest of circuit matches the supplied voltage E.
By solving the equivalent resistance in each branch and applying KVL (Kirchhoff’s Voltage Law), the total voltage required to maintain 0.5 A through the 5 Ω resistor (and rest of the configuration) is:
